In a previous question of mine, I asked how the volume of the region $$xy>zw \quad\wedge \quad x>-y$$ bounded by the unit $4$-ball (so $x^2+y^2+z^2+w^2<1$) could be calculated.

The answer to my question implied that the answer ($1/4$ the volume of the unit $4$-ball) could easily be seen through symmetry, but I never really understood the arguments, hence this question.

I'd really appreciate either a new perspective on this, or an explanation of the answerers argument, which roughly goes as follows:

  1. The condition $0<x+y$ divides the circle (the area of the projection of the total region on the $x,y$ plane, given the values of $z$ and $w$) in half (on the $x,y$ plane).
  2. Folding the negative $z$ half plane onto the positive one, to get the complementary superposition, divides the volume in half (on the $z,w$ plane).

The total result is to divide the whole ball by $4$.

I believe I understand the first part, but the second part eludes me; I cannot seem to visualize this "folding".

I think the idea is to say that for every pair of values of $(z,w)$ and $(-z,w),$ the sum of the areas of the projections of the region in question onto the $x,y$ plane for these two pairs ($(z,w)$ and $(-z,w)$) is a quarter of the area of the circle, or something like that. The answerer writes "the sum of the two areas, for the same $|z \cdot w|$ value, totals that of the semicircle." But I can't see why this should be true from the symmetries of the region.

Thank you.


Hint Each point on the $4$-ball is in precisely one of the subsets $$M_+ = \{xy > zw\}, \qquad M_- = \{xy < zw\}, \qquad M_0 = \{xy = zw\}$$ of the $4$-ball, and, being Zariski closed (and not the whole space), $M_0$ does not contribute to the volume of the region. On the other hand, the volume-preserving map $\phi : (x, y, z, w) \mapsto (-x, y, -z, w)$ interchanges $M_+$ and $M_-$.

One can readily modify this argument to treat the condition $x > -y$ simultaneously.

  • $\begingroup$ So the point of the map being volume-preserving is that $\mathrm{vol}(M_+)=\mathrm{vol}(M_+)=\frac{1}{2}V_{\text{ball}},$ right? Also, I'm feeling a bit slow today - would you be so kind as to include the modification? Thank you. $\endgroup$ – Bobson Dugnutt Jun 12 '17 at 14:58
  • $\begingroup$ I meant $\mathrm{vol}(M_+)=\mathrm{vol}(M_-)$, of course. $\endgroup$ – Bobson Dugnutt Jun 12 '17 at 15:20
  • $\begingroup$ Yes, that's right, they have the same volume, therefore both are $\frac{1}{2} V_{\mathrm{ball}}$. In the same vein, we can define $9$ subsets according to which relations ($>, =, <$) hold for each of the inequalities. This time, five of them (precisely those involving $=$) will not contribute to the volume. Now, define the volume-preserving map $\psi : (x, y, z, w) \mapsto (-x, -y, z, w)$, and show that you can map the given region to any of the other three contributing subsets with appropriate compositions of $\phi, \psi$. $\endgroup$ – Travis Willse Jun 12 '17 at 16:07
  • $\begingroup$ Thanks, that was very clear! But I'm still not sure I understand why the subsets with an $=$ sign have zero volume. I guess it is like if we had a $3$-ball $x^2+y^2+z^2\leq 1$ .. if we now change the $\leq$ to an $=$, the resulting region is the unit $2$-sphere, which is two-dimensional and therefore does not contribute to the volume of the sphere. But if the ball is 4D, the sphere becomes 3D - surely that has a volume? I think my concept of "volume" is too tied to 3D structures, but where exactly am I going awry? $\endgroup$ – Bobson Dugnutt Jun 12 '17 at 16:54
  • $\begingroup$ That's right---we're measuring $4$-dimensional volume, and, roughly speaking, $3$-dimensional objects have zero $4$-dimensional volume, in analogy with your $3$- and $2$-dimensional example. $\endgroup$ – Travis Willse Jun 12 '17 at 16:56

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